树的子结构

题目描述：

• 输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）

算法1：

• 使用先序遍历生成两棵树的序列，之后只需要判断序列是否存在即可。

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def pretravel(self, proot, list):
        if proot is None:
            return
        list.append(proot.val)
        self.pretravel(proot.left, list)
        self.pretravel(proot.right, list)
    def HasSubtree(self, pRoot1, pRoot2):
        list1 = []
        list2 = []
        self.pretravel(pRoot1, list1)
        self.pretravel(pRoot2, list2)
        if len(list1) == 0 or len(list2) == 0:
            return False
        for i in range(len(list1) - len(list2) + 1):
            if list1[i:len(list2) + i] == list2:
                return True
        return False

算法2：

• 递归查找

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def SearchSimiler(self, p1, p2):
        # B树为空，则存在
        if p2 is None:
            return True
        if p1 is None:
            return p1 == p2
        flag = False
        if p1.val == p2.val:
            # 查找该节点的左右节点
            flag = self.SearchSimiler(p1.left, p2.left) and self.SearchSimiler(p1.right, p2.right)
        # 如果flag为True，返回True，否则继续从左节点开始查找，如果还失败，从右节点开始查找
        return flag or self.SearchSimiler(p1.left, p2) or self.SearchSimiler(p1.right, p2)
    def HasSubtree(self, pRoot1, pRoot2):
        """
        :param pRoot1: TreeNode
        :param pRoot2: TreeNode
        :return:
        """
        # write code here
        if pRoot1 is None or pRoot2 is None:
            return False
        return self.SearchSimiler(pRoot1, pRoot2)